M. Achille Cazin on Internal Work in Gases. 



287 



p 2 =0*62 atmos. (experiment gave 0*60), 



6 = 282°«56, 

 Tj-0 =0°*44, 



p' =760-11, 

 jt?— // =1*2 millim. 



Example III. — Replace the reservoir B of 0*41185 cub. m. by 

 another of 0*7385 cub. m., as in the experiments. Then the mass 

 m i -\-m 2 of the gas which fills all the apparatus under the ordi- 

 nary pressure will be 



0*73850 + 0*10871 - ,_,, 

 m \+ m ^ 0*41185 + 0*10871 =1^787 kilogramme. 



Take, as in the first example, 



m x = 0*80442 kilogramme ; 



then the reservoir A contains the gas under the pressure of 3*8 

 atmospheres. 



Then 



m 2 = 1*62787-0*80442 =0*82345, 

 v 2 =0*89684 cub. m., 



fl 



p 2 =0*58 atmosphere, 

 6 =281°*74, 

 -0 = 1°*26, 

 pi =757*88 millims., 

 p—p' = 3*42 millims. 

 These values agree with the observations made in § X. of the 

 First Part. 



In order to facilitate the comparison, I have collected all the 

 numbers calculated into one Table : — 













P-P', 



v r 



v 2 . 



. Pv 



P2- 



Ti-r. 



in sulphuric 

 acid. 



cub. m. 



cub. m. 



atmos. 



atmos. 





millims. 



010871 



041185 



4-7 



000 



300 



65 



010871 



0-41185 



3-8 



0-24 



1-75 



37 



010871 



0-41185 



2-425 



0-62 



0-44 



9 



0-10871 



073850 



3-8 



0-58 



1-26 



27 



The calculations show what is the share of internal work in the 

 changes which the curve X X undergoes when the pressure p x 

 and the dimensions of reservoir B are changed. Since they in- 

 dicate exactly the direction and, up to a certain point, the mag- 

 nitude of these changes, just as they have been observed, they 

 furnish a new proof of this internal work. 



